Video Lecture
Theory For Notes Making
Gauss Theorem
The statement of the Gauss’s law may be written as follows :
The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0. In symbols,
\displaystyle \oint\limits_{{surface}}{{\overrightarrow{E}\,\cdot \,\overrightarrow{{dS}}}}\,\,=\,\,\frac{{{{q}_{{in}}}}}{{{{\varepsilon }_{0}}}} …(i)
where q is the net charge enclosed by the surface through which the flux is calculated.
It should be carefully noted that the electric field on the left-hand side of equation (i) is the resultant electric field due to all the charges existing in the space, whereas, the charge appearing on the right-hand side includes only those which are inside the closed surface.
For example for the situation show in diagram the Guass’s law can be written as
\displaystyle \oint{{(\overrightarrow{{{{E}_{1}}}}+}}\overrightarrow{{{{E}_{2}}}}+\overrightarrow{{{{E}_{3}}}}+\overrightarrow{{{{E}_{4}}}}+\overrightarrow{{{{E}_{5}}}}+\overrightarrow{{{{E}_{6}}}})\,\,\cdot \,\,\overrightarrow{{dS}}\,\,\,=\,\,\frac{{{{q}_{1}}+{{q}_{2}}+{{q}_{3}}}}{{{{\varepsilon }_{0}}}}
What is the total flux produced by a charge?
Consider a charge q . Construct a spherical surface around it with charge at the center. Now the flux passing through the spherical surface is same as that produced by the charge.
The flux passing through the spherical surface can be calculated using the equation \displaystyle \phi =EA\cos \theta .
hereE is the electric field produced by the charge hence \displaystyle E=\frac{{kq}}{{{{r}^{2}}}},
A is the surface area of the sphere hence \displaystyle A=4\pi {{r}^{2}}
andq =0 as the electric field is radial hence it is along the normal to the surface.
So \displaystyle \phi =\frac{{kq}}{{{{r}^{2}}}}.4\pi {{r}^{2}}.\cos (0)=kq4\pi
\displaystyle \Rightarrow \phi =\frac{1}{{4\pi {{\varepsilon }_{o}}}}q4\pi
hence \displaystyle \phi =\frac{q}{{{{\varepsilon }_{o}}}} This is the flux passing through the spherical surface and actually it is produced by the charge q place at the center. So we conclude that a charge q produce a total flux equal to \displaystyle \frac{q}{{{{\varepsilon }_{o}}}}.
Hence if a charge is surrounded by a number of closed surfaces of any shape as shown then the flux passing through every surface is equal to that produced by the charge i.e. \displaystyle \frac{q}{{{{\varepsilon }_{o}}}}.
Application of Gauss’s Law
Using Gauss’s law in some cases with symmetry arguments, we can derive several important results in electrostatic situations. (For detailed derivation please see the video lectures)
1.
The electric field at any point due to an infinite line of charge with uniform linear charge density λ is perpendicular to the line of charge and has magnitude
E=\frac{\lambda }{{2\pi {{\varepsilon }_{o}}r}} (line of charge)
wherer is the perpendicular distance from the line of charge to the point.
2.
The electric field due to an infinite nonconducting sheet with uniform surface charge density σ is perpendicular to the plane of the sheet and has magnitude
E=\frac{\sigma }{{2{{\varepsilon }_{o}}}} (sheet of charge)
Note that this electric field does not depend on the distance r from the sheet,hence it is a constant.
3.
The electric field outside a spherical shell of charge with radius R and total charge q is directed radially and has magnitude
E=\frac{1}{{4\pi {{\varepsilon }_{o}}}}\frac{q}{{{{r}^{2}}}} (spherical shell, for r ≥ R).
Here r is the distance from the center of the shell to the point at which E is measured. (The charge behaves, for external points, as if it were all located at the center of the sphere). The field inside a uniform spherical shell of charge is exactly zero.
E=0 (spherical shell, for r<R).
4.
The electric field inside a uniformly charged non conducting sphere is directed radially and has magnitude
E=\left( {\frac{q}{{4\pi {{\varepsilon }_{o}}{{R}^{3}}}}} \right)r=\frac{{\rho r}}{{3{{\varepsilon }_{o}}}}
whereas outside it is given as E=\frac{1}{{4\pi {{\varepsilon }_{o}}}}\frac{q}{{{{r}^{2}}}}
Illustration
The electric field of two point charges separated by 4.14 cm is as shown in figure (a) What is the nature of charges? (b) What is the ratio of magnitude of charges? (c) Is the field uniform ? (d) Apart from infinity where is the neutral point?
Solution
(a) As lines of force are starting from A and terminating on B, A is positive charge while B negative charge
(b) As number of lines of force emerging or terminating on a charge is proportional to its magnitude.
\displaystyle \left| {\frac{{{{q}_{A}}}}{{{{q}_{B}}}}} \right|\,\,=\,\,\frac{{12}}{6}\,\,=\,\,\frac{2}{1}\, i.e. \displaystyle \left| {{{q}_{A}}} \right|\,\,=\,\,2|\,\,{{q}_{B}}\,|\,
(c) As the lines of force are not equidistant straight lines, the field is not uniform.
(d) As charge \displaystyle {{q}_{A}} is positive while \displaystyle {{q}_{B}} negative neutral point cannot be between them. So if neutral point is at a distance x from B on the line AB.
them. So if neutral point is at a distance x from B on the line AB.
\displaystyle \frac{{{{q}_{A}}}}{{{{{(d+x)}}^{2}}}}\,\,+\,\,\frac{{{{q}_{B}}}}{{{{x}^{2}}}}\,\,=\,\,0 or \displaystyle \frac{{2{{q}_{B}}}}{{{{{(4.14+x)}}^{2}}}}\,\,=\,\,\frac{{{{q}_{B}}}}{{{{x}^{2}}}}
Or \displaystyle \sqrt{2}\,\,(x)\,\,=\,\,(4.14\,+\,x) or x = 10 cm
i.e., neutral point is at C such that BC = 10 cm
Illustration
A spherical shell of radius a is placed in a uniform electric field \displaystyle \vec{E}. Find the amount of flux entering into the shell
Solution
It is the shaded circular area or projected area of shell which is intercepting the flux hence flux entering depends on this area
Φ = E . S = E .πa^2
Illustration
Derive Gauss’ law from Coulomb’s law.
Solution
Let us consider a point charge q inside an arbitrary closed surface S.
The electric field due to q at the position of a differential element \displaystyle d\overrightarrow{S} is given by
\displaystyle \overrightarrow{E}=\frac{q}{{4\pi {{\in }_{0}}{{r}^{2}}}}\hat{e}
where \displaystyle \hat{e} is the unit vector pointing radially outwards from the charge
The flux through \displaystyle d\overrightarrow{S} is given by
\displaystyle d\Phi =\overrightarrow{E}.d\overrightarrow{S}=\frac{q}{{4\pi {{\in }_{0}}{{r}^{2}}}}\hat{e}.d\overrightarrow{S}=\frac{q}{{4\pi {{\in }_{0}}}}\frac{{dS\cos \theta }}{{{{r}^{2}}}}=\frac{q}{{4\pi {{\in }_{0}}}}d\Omega
Thus, \displaystyle \Phi =\int{{d\Phi }}=\frac{q}{{4\pi {{\in }_{0}}}}\int{{d\Omega }}=\frac{I}{{4\pi {{\in }_{0}}}}[4\pi ]\,\,=\,\,\frac{q}{{{{\in }_{0}}}}
Illustration
Determine the field at a point distant r from a uniformly charged plane with surface charge density σ.
Solution
By symmetry, we conclude the \displaystyle \overrightarrow{E} can only be normal to the charged plane. Also, at points A and B, symmetric with respect to the plane, \displaystyle \overrightarrow{E} should have same magnitude but opposite directions. Hence, a right cylinder is chosen as Gaussian surface.
The flux through the lateral surface is zero. If ΔS is the area of each end face,the total flux is
ΔΦ = EΔS + EΔS = 2EΔS.
Then according to Gauss’s Law,
2EΔS = \frac{{{{q}_{{in}}}}}{{{{\varepsilon }_{0}}}}\text{ }=\text{ }\frac{{\sigma \Delta S}}{{{{\varepsilon }_{0}}}}
\displaystyle E=\frac{\sigma }{{2{{\varepsilon }_{0}}}}
Illustration
A uniform positive charge density r (charge per unit volume) exists throughout a spherical volume of radius R. Find the electric field (a) outside the sphere and (b) inside the sphere. Also show how electric field varies as the distance of the point from the centre increases from zero to infinite.
Solution
By symmetry, we conclude that the field \displaystyle \overrightarrow{E} can only be radially outward, both inside and outside the sphere. Moreover, for a given value of r, \displaystyle \overrightarrow{E} has the same magnitude everywhere. To match this symmetry, we choose a Gaussian surface in the form of a sphere of radius r, centred on the spherical volume.
(a)
For any point outside the sphere, r > R :
The total charge inside the spherical volume (surface A) is
\displaystyle Q=\int{{\rho dV}}=\rho \left( {\frac{4}{3}\pi {{R}^{3}}} \right) …(i)
Applying Gauss’s Law,
\oint{{\vec{E}.d\vec{S}}} = \frac{{{{q}_{{\text{in}}}}}}{{{{\varepsilon }_{0}}}}
\displaystyle E(4\pi {{r}^{2}})=\frac{Q}{{{{\varepsilon }_{0}}}}
or \displaystyle E=\frac{Q}{{4\pi {{\varepsilon }_{0}}{{r}^{2}}}} …(ii)
This is just the inverse-square-law field for a point charge Q. In fact, this result is valid for any charged sphere, whether solid or hollow, conducting or non-conducting. The sphere behaves as if whole charge is concentrated at its centre. We can express the result of Eqn.(ii) in terms of the given parameters by using Eqn. (i),
\displaystyle E=\frac{Q}{{4\text{ }\pi {{\varepsilon }_{0}}{{r}^{2}}}}=\frac{1}{{4\text{ }\pi {{\varepsilon }_{0}}{{r}^{2}}}} \rho \left( {\frac{4}{3}\pi {{R}^{3}}} \right)=\frac{{\rho {{R}^{3}}}}{{3\text{ }{{\varepsilon }_{0}}}} \frac{1}{{{{r}^{2}}}}
i.e. E ∝ \displaystyle \frac{1}{{{{r}^{2}}}}
(b)
For any point inside the sphere (r<R) :
We now choose a Gaussian surface B, so that r<R. The charge inside this sphere is
\displaystyle {{q}_{{in}}} = \int\limits_{0}^{r}{{\rho dV}} = \rho \left( {4\text{/3 }\pi {{r}^{3}}} \right)
Applying Gauss’s law,
\displaystyle \oint{{\overrightarrow{E}\,\,.\,\,d\overrightarrow{S}}} = \frac{{{{q}_{{in}}}}}{{{{\varepsilon }_{0}}}}
or \displaystyle E(4\text{ }\pi \text{ }{{r}^{2}})\text{ }=\text{ }\frac{{4\text{ }\rho \text{ }\pi \text{ }{{r}^{3}}}}{{3{{\varepsilon }_{0}}}}
or E = \displaystyle \frac{\rho }{{3{{\varepsilon }_{0}}}}r
i.e. E ∝ r
Thus, inside the sphere of uniform charge, the field is directly proportional to the distance r from the centre. The variation of E with distance r is shown in the figure.
Illustration
Determine and draw the graph of electric field due to infinitely large non conducting sheet of thickness t and uniform volume charge density σ as a function of distance x from its symmetry plane for .
(a) \displaystyle x\le \frac{t}{2}
(b) \displaystyle x\ge \frac{t}{2}
Solution
We can assume thick sheet to be made of large number of uniformly charged thin sheets. Consider an elementary thin sheet of width dx at a distance x from symmetry plane.
Charge in sheet = ρAdx
(A : assumed area of sheet)
surface charge density
\sigma =\frac{{\rho Adx}}{A}
so, electric field intensity due to elementary sheet. dE=\frac{{\rho dx}}{{2{{\varepsilon }_{0}}}}
(a)
When x < t/2
{{E}_{{net}}}=\int\limits_{{-t/2}}^{x}{{\frac{{\rho dx}}{{2{{\varepsilon }_{0}}}}}}-\int\limits_{x}^{{t/2}}{{\frac{{\rho dx}}{{2{{\varepsilon }_{0}}}}}}=\frac{{\rho x}}{{{{\varepsilon }_{0}}}}
(b)
When x > t/2
{{E}_{{net}}}=\int\limits_{{-t/2}}^{{t/2}}{{\frac{{\rho dx}}{{2{{\varepsilon }_{0}}}}}}=\frac{{\rho t}}{{2{{\varepsilon }_{0}}}}
Illustration
Two conducting plates of large surface area M and N are placed parallel to each other. M is given a charge \displaystyle {{Q}_{1}} and N is given a charge \displaystyle {{Q}_{2}}. Find the distribution of charges on the four surfaces.
Solution
Consider the closed Gaussian surface as indicated by the dashed line. The flux through two faces which lie inside the plates is zero because there is no electric field inside the conductors. The flux through other four surface is zero because electric field lines do not pass through there surfaces.
Therefore, the electric flux through the closed surface is zero. From gauss’s law, the total charge enclosed should be zero. Hence, the charge on two opposite faces should be equal and opposite. The distribution is shown in the other figure.
To find q, we consider a point P inside plate N. Let the area of the plates be A.
Using formula E=\frac{\sigma }{{2{{\varepsilon }_{o}}}}
the electric field at P due to
charge Q1-q = \frac{{{{Q}_{1}}-q}}{{2A{{\varepsilon }_{0}}}} ( rightward)
charge + q = \frac{q}{{2A{{\varepsilon }_{0}}}} ( rightward)
charge – q = \frac{q}{{2A{{\varepsilon }_{0}}}} ( leftward)
charge Q2+q = \frac{{{{Q}_{2}}+q}}{{2A{{\varepsilon }_{0}}}} ( leftward)
E=\frac{{{{Q}_{1}}-q}}{{2A{{\varepsilon }_{0}}}}+\frac{q}{{2A{{\varepsilon }_{0}}}}-\frac{q}{{2A{{\varepsilon }_{0}}}}-\frac{{{{Q}_{2}}+q}}{{2A{{\varepsilon }_{0}}}}
Since point P is inside the conductor, this field should be zero.
Therefore Q1-q = Q2+q or q=\frac{{{{Q}_{1}}+{{Q}_{2}}}}{2}
Therefore the charge distribution is as shown in the figure below.
Objective Assignment
Q.1
A charge Q is located at the centre of a cube. The electric flux through any face of the cube is
(a) Q/12 eo
(b) 4p Q/eo
(c) Q/6 eo
(d) Q/4 peo
Ans. (c)
Q.2
How many electrons must be added to a spherical conductor of radius 10 cm to produce a field of 2 ´ 10–3 N/C just above the surface
(a) 1.39 ´ 104
(b) 1.6 ´ 105
(c) 3 ´ 104
(d) 9.1 ´ 104
Ans. (a)
Q.3
A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by
(a) 2pR2E
(b) 2pR2/E
(c) zero
(d) none of these
Ans. (c)
Q.4
A hemispherical surface of radius R is placed with its cross-section perpendicular to a uniform electric field E as shown in figure. Flux linked with its curved surface is
(a) zero
(b) 2\pi {{R}^{2}}E
(c) pR2E
(d) \frac{E}{{2{{\varepsilon }_{o}}}}
Ans. (c)
Q.5
A hollow conducting sphere of charge does not produce an electric field at any
(1) Interior point
(2) Outer point
(3) Point beyond 2 metre
(4) Point beyond 10 metre
Ans. (1)
Q.6
A positive point charge, which is free to move, is placed inside a hollow conducting sphere, placed away from it centre. The conducting sphere is uniformly charged with negative charge. The positive charge will
(1) Move towards the centre
(2) Move towards the nearer wall of the conductor
(3) Remain stationary
(4) Oscillate between the centre and the nearer wall
Ans. (3)
Q.7
How does the electric field strength vary when we enter a uniformly charged spherical cloud ?
(1) Decreases inversely as the square of the distance from the surface.
(2) Decreases directly as the square of the distance from the surface
(3) Increase directly as the square of the distance from the centre
(4) Increases directly as the distance from the centre.
Ans. (4)
Q.8
Gauss’s law helps in
(1) Determination of electric force between point charges
(2) Situations where Coulomb’s law fails
(3) Determination of electric field due to symmetric charge distributions
(4) Determining electric potential due to symmetric charge distributions.
Ans. (3)
Q.9
Which one of the following graphs, figures shows the variation of electric field strength E with distance d from the centre of the hollow conducting sphere ?
Ans (4)
10.
If an insulated non-conducting sphere of radius R has charge density r. The electric field at a distance r from the centre of sphere (r<R) will be
(a) \frac{{\rho R}}{{3{{\varepsilon }_{0}}}}
(b) \frac{{\rho r}}{{{{\varepsilon }_{0}}}}
(c) \frac{{\rho r}}{{3{{\varepsilon }_{0}}}}
(d) \frac{{3\rho R}}{{{{\varepsilon }_{0}}}}
Ans (c)
11.
Shown in the figure is a distribution of charges. The flux of electric field due to these charges through the surface is
(a) 3 q / e0
(b) 2 q / e0
(c) q / e0
(d) Zero
Ans (d)
12.
The inward and outward electric flux for a closed surface in units of N-m2/coulomb are respectively 8 × 103 and 4 × 103. Then the total charge inside the surface is
(a) 4 × 103 coulomb
(b) – 4 × 103 coulomb
(c) -\frac{{4\times {{{10}}^{3}}}}{{{{\varepsilon }_{0}}}}coulomb
(d) – (4 × 103 )e0 coulomb
Ans (d)
Subjective Assignment
Q.1
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube ?
Q.2
An electric dipole of dipole moment \displaystyle 20\,\,\times \,{{10}^{{-6}}} cm is enclosed by a closed surface. What is the net flux coming out of the surface ?
Q.3
Is the electric field due to a charge configuration with total charge zero, necessarily zero ? Justify.
Q.4
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Q.5
An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
Q.6
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 \displaystyle \mu \,C/{{m}^{2}}.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere ?
Q.7
A spherical conducting shell of inner radius \displaystyle {{r}_{1}} and outer radius \displaystyle {{r}_{2}} has a charge ‘Q’. A charge ‘q’ is placed at the centre of the shell.
(a) What is the surface charge density on the (i) inner surface, (ii) outer surface of the shell ?
(b) Write the expression for the electric field at a point \displaystyle x\,>\,{{x}_{2}} from the centre of the shell.
Q.8
A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Q.9
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.
Q.10
State Gauss’s law in electrostatics. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet.
Q.11
Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density \displaystyle \lambda C{{m}^{{-1}}}.
Q.12
Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities and – respectively.
Q.13
Using Gauss’s law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r>R and r < R.
Q.14
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 Nm2/C.
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?
Q.15
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Q.16
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Q.17
State Gauss’s Law in electrostatics. Consider an overall neutral sphere of radius R. This sphere has a point charge, +Q, at its centre and this positive charge is surrounded by a uniform density of negative charge up to a radius R. Use Gauss’s law to obtain expressions for the electric field, of this sphere, at a point distant r, from its centre, where (i) r < R (ii) r > R. Show that these two expressions give identical results, for the electric field at r = R.
Q.18
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
